Cho : S=1+1/(2^1/2)+1/(3^1/2)+...+1/(100^1/2)
CMR:18<S<19
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Cho \(S=1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{100}}\). CMR: \(18< S< 19\)
\(S=\frac{2}{2}+\frac{2}{2\sqrt{2}}+\frac{2}{2\sqrt{3}}+...+\frac{2}{\sqrt{100}}\)
\(S>\frac{2}{\sqrt{1}+\sqrt{2}}+\frac{2}{\sqrt{2}+\sqrt{3}}+\frac{2}{\sqrt{3}+\sqrt{4}}+...+\frac{2}{\sqrt{100}+\sqrt{101}}\)
\(S>2\left(\sqrt{2}-\sqrt{1}\right)+2\left(\sqrt{3}-\sqrt{2}\right)+...+2\left(\sqrt{101}-\sqrt{100}\right)\)
\(S>2\left(\sqrt{101}-1\right)>2\left(\sqrt{100}-1\right)=18\) (1)
\(S< 1+\frac{2}{\sqrt{1}+\sqrt{2}}+\frac{2}{\sqrt{2}+\sqrt{3}}+...+\frac{2}{\sqrt{99}+\sqrt{100}}\)
\(S< 1+2\left(\sqrt{2}-1\right)+2\left(\sqrt{3}-\sqrt{2}\right)+...+2\left(\sqrt{100}-\sqrt{99}\right)\)
\(S< 1+2\left(\sqrt{100}-1\right)=19\) (2)
(1); (2) \(\Rightarrow18< S< 19\)
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CMR \(2\left(\sqrt{n+1}-\sqrt{n}\right)< \frac{1}{\sqrt{n}}< 2\left(\sqrt{n}-\sqrt{n-1}\right)\)với n thuộc N*
Áp dụng cho S=\(1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{100}}\)
CMR 18<S<19
S=1+1/2^2+1/3^2+...+1/100^2
CMR S<2
Câu 2: CMR S<1/4 với S=1/4^2+1/6^2+...+1/(2n)^2
\(S=1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{100^2}< 1+\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}\)
Mà \(1+\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}=1+1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(=2-\dfrac{1}{100}< 2\)
\(\Rightarrow\) \(S< 1+\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}\)
Vậy \(S< 2\left(đpcm\right).\)
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Câu 1 :
Ta có :
\(S=1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+..........+\dfrac{1}{100^2}\)
Ta thấy :
\(\dfrac{1}{2^2}< \dfrac{1}{1.2}\)
\(\dfrac{1}{3^2}< \dfrac{1}{2.3}\)
........................
\(\dfrac{1}{100^2}< \dfrac{1}{99.100}\)
\(\Leftrightarrow S< 1+\dfrac{1}{1.2}+\dfrac{1}{2.3}+.......+\dfrac{1}{99.100}\)
\(\Leftrightarrow S< 1+1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+.....+\dfrac{1}{99}-\dfrac{1}{100}\)
\(\Leftrightarrow S< 1+1-\dfrac{1}{100}\)
\(\Leftrightarrow S< 2+\dfrac{1}{100}< 2\)
\(\Leftrightarrow S< 2\rightarrowđpcm\)
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\(S=1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{100^2}\)
\(S< 1+\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}\)
\(S< 1+1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(S< 2-\dfrac{1}{100}\)
\(S< 2\rightarrowđpcm\)
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Xem thêm câu trả lời
S=1+1/2+1/3+.....+1/2100-1
CMR: S<100
S=1/20+(1/21+1/22-1)+(1/22+...+1/23-1)+...+(1/299+...+1/2100-1) (100 cặp)
S<1/20.20+1/21.21+1/22.22+...+1/299.299
S<1+1+1+...+1 (100 số 1)
S<100.1
S<100 (ĐPCM)
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Cmr: S= 1+1/2^2+1/3^2+1/4^2+...+1/100^2>3/2
Cho `S=1/(5^2) + 2/(5^3) + 3/(5^4) + ... + 99/(5^100)` CMR `S<1/16`
Ta có :
`5S=5(1/(5^2)+2/(5^3)+3/(5^4)+...+99/(5^100))`
`5S=1/5+2/(5^2)+3/(5^3)+...+99/(5^100)`
`=>5S-S=1/5+2/(5^2)+3/(5^3)+...+99/(5^100)-(1/(5^2)+2/(5^3)+3/(5^4)+...+99/(5^100))`
`4S=1/5+1/(5^2)+1/(5^3)+1/(5^4)+...+1/(5^99) -99/(5^100)`
`20S=5(1/5+1/(5^2)+1/(5^3)+...+1/(5^99)-99/(5^100))`
`20S=1+1/5+1/(5^2)+....+1/(5^98)-99/(5^99)`
`=>20S-4S=(1+1/5+1/(5^2)+...+1/(5^98)-99/(5^99))-(1/5+1/(5^2)+1/(5^3)+...+1/(5^99)-99/(5^100))`
`=>16S=1-99/(5^99)-1/(5^99)-99/(5^100)`
Vì `-99/(5^99)-1/(5^99)-99/(5^100)<0=>1-99/(5^99)-1/(5^99)-99/(5^100)<1`
`=>S<1/16`
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CMR : S < 3 biết : 1 + 1/2^2 + 1/3^2 + 1/4^2 + ... + 1/100^2
\(2\left(\sqrt{N+1}-\sqrt{N}\right)<\frac{1}{\sqrt{N}}<2\left(\sqrt{N}-\sqrt{N-1}\right)\)
Với N>0
Áp dụng: cho s=\(1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{100}}\)
Cmr 18<s<19
1, Cho S=17+172+173+....+1718
a, CMR S chia hết cho 307
b, Tìm ƯCLN(S,18)
2, Cho B= 7+73+75+77+.....799
a, Tính B
b, CMR (7100-1) cHIa HẾT Cho 48